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aceace
 2 years ago
Best ResponseYou've already chosen the best response.0Find a relation between x and y that does NOT involve logarithms: log x + log y = log (x+ y)

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3Can we assume it's a log of base 10?

AERONIK
 2 years ago
Best ResponseYou've already chosen the best response.0logx + log y is never log(x+y) it seems... it is log(xy)

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3I suppose we could exponentiate both sides.. \[\huge 10^{(\log x + \log y)}=10^{\log(x+y)}\] Using rules of exponents we can split up the left side like this: \[\huge 10^{\log x}*10^{\log y}=10^{\log(x+y)}\] Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". Giving us: \[\huge xy=(x+y)\] \[\huge y=\frac{x}{x1}\]

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3Something like that maybe? :o

aceace
 2 years ago
Best ResponseYou've already chosen the best response.0@zeptr yes thats the answer !!!

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3Were you just testing us or something? :D lol Knew the answer the whole time huh? XD

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3@AERONIK , it seems that logx+logy=log(x+y) sometimes. I wouldn't say NEVER. Obviously it doesn't match the rule for logs that we had in mind. But if you plug in y=2, x=2 into the original equation, that should work out ok as one solution :D Since 2+2 is the same as 2*2

aceace
 2 years ago
Best ResponseYou've already chosen the best response.0no i had the answers at the back of book but i have no idea how to do it... i still dont get the logic though...

AERONIK
 2 years ago
Best ResponseYou've already chosen the best response.0so it is not an identity and just a function you say ?

aceace
 2 years ago
Best ResponseYou've already chosen the best response.0yeh just a random equation

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3Yah it appears it wasn't testing you on whether or not you knew your log identities, they just wanted to see a relationship between x and y. Interesting problem :O

AERONIK
 2 years ago
Best ResponseYou've already chosen the best response.0and one more thing in your proof you have assumed the given statement to be correct, can you please give a formal proof for te same?

AERONIK
 2 years ago
Best ResponseYou've already chosen the best response.0and is it log to the base 10 or log to the base e ?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3For this problem, we have to assume that they are ALL the same base. Otherwise it won't work out the same. If they're a different base, let's say A, then we exponentiate, writing each side with a base A (regardless of what A might be) and it will work out the same! :D

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3Different than 10 i mean*

aceace
 2 years ago
Best ResponseYou've already chosen the best response.010 logx ∗10 logy =10 log(x+y) Recognizing that log base 10, and the exponentiation base 10 are inverse operations of one another, they essentially "cancel out". HOW DID YOU DO THIS STEP... CAN YOU SHOW ME? PLEASE

AERONIK
 2 years ago
Best ResponseYou've already chosen the best response.0by defination if y=log x to the base a, then x is nothing but a^y, just use this and you can prove the identity.

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3\[\large \arcsin( \sin x)=x\]\[\huge e^{\ln x}=x\] Yah this is true of any function and it's inverse :D Remember it in trig with the inverse functions?

aceace
 2 years ago
Best ResponseYou've already chosen the best response.0no ...can you do a proof for it please?

zepdrix
 2 years ago
Best ResponseYou've already chosen the best response.3dw:1353225574033:dw Hmmm this might be a little confusing.. I drew a lot of arrows : lemme know if it makes some sense.

aceace
 2 years ago
Best ResponseYou've already chosen the best response.0i get it ...thanks so much
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